The length of a rectangle is 3 inches more than twice its width, and its area is 65 square inches. What is the width?If w = the width of the rectangle, which of the following expressions represents the length of the rectangle?
2w + 3
2(w + 3)
3(2w)
Answers
65=2W^2+3W
2W^2+3W-65=0
We have a quadratic equation, hence, we will solve for its factors. We have,
(2w+13)(w-5)=0
The two roots of the equation are: -13/2 and 5. Since we can only have positive numbers for dimensions, the width of the rectangle is 5 in.
The expression for the length of the rectangle is 2W+3.
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Answers
our variable can me L for Length
3L+4L+5L=90
now all we have to do is solve
12L=90
---- ----
12 12
L=7.5
now multiple each number in the ratio by 7.5 to find out how long each side is
3 times 7.5 is 22.5
4 times 7.5 is 30
5 times 7.5 is 37.5
the side lengths are 22.5, 30, and 37.5
A. 4/3
B. 2/3
C. 7/9
D. 4/27
Answers
Answers
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Explanation:
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m∡7 = m∡1 ; SInce these angles are vertical; and vertical angles are congruent.
m∡1 + m∡2 = 180 ; since these angles are supplementary; and by definition; supplementary angles "add up" to 180.
Given: m∡2 = 123 ;
m∡1 + 123 = 180 ;
Since m∡1 = m∡7 ; and we wish to find: "m∡7" ; we can rewrite as:
_______________________________
m∡7 + 123 = 180 ; solve for "m∡7" ;
_______________________________
Subtract "123" from EACH SIDE of the equation:
_________________________________________
m∡7 + 123 - 123 = 180 - 123 ;
_________________________________________
m∡7 = 57° .
_________________________________________
122√
63√
6
A right triangle with both legs shown with one tick mark. One of the legs is labeled 12. The hypotenuse is labeled x.
Answers
Here we use the Pythagoras theorem
According to this, in a right angled triangle, the sum of the squares of the lengths of the two sides is equal to the hypotenuse
is "a" and "b" are the sides, and "h" is the hypotenuse, then :
In the given problem, a = b = 12 and h = x
12^2 + 12^2 = x^2
So, x =
x =
X =
X = 12 (Option B)
1. 2x+3y=1
-3x+y=15
Answers
Answers
Answer:
The length and width that maximize the area are:
W = 2*√8
L = 2*√8
Step-by-step explanation:
We want to find the largest area of a rectangle inscribed in a semicircle of radius 4.
Remember that the area of a rectangle of length L and width W, is:
A = L*W
You can see the image below to see how i will define the length and the width:
L = 2*x'
W = 2*y'
Where we have the relation:
4 = √(x'^2 + y'^2)
16 = x'^2 + y'^2
Now we can isolate one of the variables, for example, x'
16 - y'^2 = x^'2
√(16 - y'^2) = x'
Then we can write:
W = 2*y'
L = 2*√(16 - y'^2)
Then the area equation is:
A = 2*y'*2*√(16 - y'^2)
A = 4*y'*√(16 - y'^2)
If A > 1, like in our case, maximizing A is the same as maximizing A^2
Then if que square both sides:
A^2 = (4*y'*√(16 - y'^2))^2
= 16*(y'^2)*(16 - y'^2)
= 16*(y'^2)*16 - 16*y'^4
= 256*(y'^2) - 16*y'^4
Now we can define:
u = y'^2
then the equation that we want to maximize is:
f(u) = 256*u - 16*u^2
to find the maximum, we need to evaluate in the zero of the derivative:
f'(u) = 256 - 2*16*u = 0
u = -256/(-2*16) = 8
Then we have:
u = y'^2 = 8
solving for y'
y' = √8
And we know that:
x' = √(16 - y'^2) = √(16 - (√8)^2) = √8
And the dimensions was:
W = 2*y' = 2*√8
L = 2*y' = 2*√8
These are the dimensions that maximize the area.